Find the integral values of x and y satisfying the inequality 3y + 5x \(\leq\) 15, given that y > 0, y < 3, and x > 0.
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( 1, 1), (1, 2), (2, 1)
- ( 1, 1), (1, 3), (2, 2)
- ( 1, 1), (1, 2) (1, 3)
- ( 1, 1), (1, 3), (2, 1)
The correct answer is: A
Explanation
From the constraint y < 3 and y > 0, the possible values for y are 1 and 2
When y = 1, 3y + 5x \(\leq\) 15
3 + 5x \(\leq\) 15 ⇒ 5x \(\leq\) 12 ⇒ 2.4
since x must be an integer greater than 0 but less than or equal to 2.4, the possible values of x are 1 and 2
When y = 2, 3y + 5x \(\leq\) 15 ⇒ 6 + 5x \(\leq\) 15 ⇒ 6 \(\leq\) 15 - 6 = 11 ⇒ x = 2.2
Again, Since x must be an integer greater than 0 and less than or equal to 2.2, the only possible integer value of x is 1
So, the pairs (x, y) that satisfy all the conditions are ( 1, 1), (1, 2) and (2, 1).
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